Question

What are vertices of the conic 16x² - 25y² = 400 ?

Answer 1

(-5, 0) and (5, 0)

Explanation:

This equation fits the form for a hyperbola with x-intercepts. The standard form for such an equation is

`(x^2)/(a^2)-(y^2)/(b^2)=1`

To get the equation in the question into this standard form, divide each term by 400.

`(16x^2)/(400)-(25y^2)/(400)=(400)/(400)\\(x^2)/(25)-(y^2)/(16)=1`

To find the x-intercepts, make y = 0.

`(x^2)/(25)=1\\x^2=25\\x=\pm 5`

The vertices are located at the points (-5, 0) and (5, 0).

Note: There are no y-intercepts; making x = 0 produces no real solutions for y.